Medium
Given an integer array nums
, return the length of the longest strictly increasing subsequence.
A subsequence is a sequence that can be derived from an array by deleting some or no elements without changing the order of the remaining elements. For example, [3,6,2,7]
is a subsequence of the array [0,3,1,6,2,2,7]
.
Example 1:
Input: nums = [10,9,2,5,3,7,101,18]
Output: 4
Explanation: The longest increasing subsequence is [2,3,7,101], therefore the length is 4.
Example 2:
Input: nums = [0,1,0,3,2,3]
Output: 4
Example 3:
Input: nums = [7,7,7,7,7,7,7]
Output: 1
Constraints:
1 <= nums.length <= 2500
-104 <= nums[i] <= 104
Follow up: Can you come up with an algorithm that runs in O(n log(n))
time complexity?
defmodule Solution do
@spec length_of_lis(nums :: [integer]) :: integer
def length_of_lis(nums) do
Enum.reduce(nums, :gb_sets.new(), fn x, gb ->
:gb_sets.iterator_from(x, gb)
|> :gb_sets.next()
|> then(fn
:none -> :gb_sets.insert(x, gb)
{^x, _} -> gb
{y, _} ->
:gb_sets.delete(y, gb)
|> then(&:gb_sets.insert(x, &1))
end)
end)
|> :gb_sets.size()
end
end